john smith wrote:
depending in the controll , use the macro funtion
and have the path reduced by the "opisite and ajasin " per pass , ya know
how we did it before "Cad-Cam" ??

"Eat at Sloppy Joes" wrote in message
...

I'm trying to mill a conical hole using CNC. I'm having trouble
programming
a helical cut using and end mill. As it's sunday I can't pick up a
tapered
reamer to fit my needs.

Here's the geometry:
0.5" thick mild steel
0.5" diameter hole on top, 0.375" diameter at bottom on XY plane.


Here's the problem:
Use a 3/8" ball mill to make this hole. The hole has been pre drilled.

Anyone have any G-Code samples to make this hole?


Your going to have to do the work on this one. It will be tedious, but
simple. (.5-.375)/2 = .0625. To figure out the angle take your side
opposite .0625 and divide it by your adjacent .5 = .125, inverse tangent
comes out to 7.125 degrees. Next you have to determine your stepover
amount SQRT((Finish i.e. 63)*(8*tnr .1875))/1000000 = .0097, we'll round
up to .01. Now calculate your side opposite (that is what you will
subtract your radius by every time you step down in Z .01, .01*TAN7.125
= .00125. That's it, now get to work. You probably should calculate your
start coordinates at about .02 above the work and your finish
coordinates .02 below the work. Here is your first circle. (.00125*2)*2
= .005, (.505-.375)/2 = .065, your first coordinates will be Z.02 and a
..065 radius, 2nd .065 - .00125 Z.01 and a radius of .06375. Good luck
and have fun.